HDU 4282 A very hard mathematic problem 2分
HDU 4282 A very hard mathematic problem 二分
来源:http://acm.hdu.edu.cn/showproblem.php?pid=4282
题意:给出一个数n,问x^z + y^z + x*y*z = n有多少这样的x y z,其中y > x,z > 1,x,y,z都是正数。
思路:注意到z最大为31,x最大为50000,因此可以枚举x,z,二分判断y即可。
代码:
#include <iostream> #include <vector> #include <cstdio> #include <string.h> #include <vector> #include <climits> using namespace std; const int N = 50010; __int64 aa[N][35]; __int64 mi(int x,int cnt){ __int64 s = 1; for(int i = 1; i <= cnt; ++i) s *= x; return s; } void init(){ memset(aa,-1,sizeof(aa)); for(int i = 1; i < N; ++i){ for(int j = 2; j <= 31; ++j){ __int64 x = mi(i,j); if(x >= INT_MAX || x < 0) continue; aa[i][j] = x; } } } int main(){ //freopen("1.txt","r",stdin); init(); int n; while(scanf("%d",&n) && n){ int ans = 0; for(int x = 1; x < N; ++x){ for(int z = 2; z <= 31; ++z){ if(aa[x][z] == -1) break; int lp = 2,rp = N - 1; bool flag = false; while(lp <= rp){ int mid = (lp + rp) / 2; if(aa[mid][z] == -1){ rp = mid - 1; continue; } __int64 sum = aa[x][z] + mi(mid,z) + x*mid*z; if(sum == n ){ if(mid > x){ flag = true; break; } rp = mid - 1; } else if(sum < n){ lp = mid + 1; } else{ rp = mid - 1; } } if(flag) ans++; } } printf("%d\n",ans); } return 0; }